Cut The Knot!

An interactive column using Java applets
by Alex Bogomolny

A Matter of Appreciation

October 1999

I have a recollection. Years ago, a childhood friend of mine, Boris, shared with me with excitement an unusual experience he had on a visit to the Tretj'yakov Art Gallery in Moscow. He was accompanied by a professional painter, a good acquaintance of his older sister. While Boris was making a round in one of the halls, he observed that the painter remained all that time on the same spot studying a certain picture. Curious, my friend asked the painter what was it about the picture that kept him interested in it for so long. According to Boris, the painter did not reply directly, but, instead, stepped over to the picture and covered a spot on the picture with a palm of his hand. "Have a look at the picture and think of what you see," he requested. After a while, he uncovered the spot, stepped back and asked Boris to have another look.

Well, almost 4 decades later, with the names of the painter and the picture long forgotten, I still vividly remember Boris' excitement when he told me of how entirely different, deeper and more beautiful, the picture appeared to him then.

This recollection is haunting me. In retrospect, I regret to have never arranged with Boris to visit the gallery and learn how to really see the picture. At the time, we were in our early teens and lived in a remote neighborhood on the outskirts of Moscow. We could not by ourselves travel half the way across the big city, and, by the time I knew how to do that, other things, quite naturally, took priority in my life.

I also remember this episode every time I go to an art exhibition. Have I had this lesson on the art of seeing pictures, would I draw more enjoyment from my visit?

I believe I would. It is possible to learn to appreciate painting without being a painter, and the point I want to make is that, by analogy, it is possible to appreciate mathematics without being a mathematician. Appreciation does not require mastery, but is impossible without adequate knowledge. Knowledge and mastery are labels at the two ends of a scale that stretches from the ability to recognize information through the ability to process it. Potential for appreciation is an increasing function on this scale. (Thus, if differentiable, the function has a positive derivative. Furthermore, appreciation feeds on itself and, through a positive feedback loop, seeks to expand knowledge. It then follows that if the appreciation function is sufficiently smooth its second derivative is also positive.)

The appreciation function depends on many parameters: disposition, perseverance, cultural surroundings, educational experience. The latter derives from mathematical subjects themselves but even more from the manner in which mathematics has been presented.

Even without a reference to the current research into knowledge representation, it's obvious that mathematics is not a collection of disparate facts. A reasonable metaphor to think of mathematics (as probably any other branch of knowledge) is a network of hierarchical structures - classes, and methods. Appreciation of mathematics grows with the discernment of links between apparently dissimilar pieces of information and the perception of the general in the particular. Facts presented in a manner that conforms to the architectural design of mathematics have a better shot at being appreciated. A worthy goal to strive to.

An elegant theorem has been published by Giovanni Ceva in 1678.

Dan Pedoe remarks in his geometry course: The theorems of Ceva and Menelaus naturally go together, since the one gives the conditions for lines through vertices of a triangle to be concurrent, and the other gives the condition for points on the sides of a triangle to be collinear.

Menelaus of Alexandria worked in the 1st century A.D., Giovanni Ceva (1648-1734) was an Italian engineer and geometer who lived some 16 centuries later. Ceva proved his theorem considering centers of gravity and the law of moments [Episodes] - the fundamental tool of Archimedes' Method (3rd century B.C.) How did it happen that the theorem remained unknown until the late 17th century? Given its simplicity, the timing of the theorem is remarkable, but also inspiring for young minds: may there still lie around simple overlooked facts waiting to be discovered?

The theorem refers to three lines AD, BE, and CF through the vertices of ΔABC. In Ceva's honor lines that connect a vertex with a point on the opposite side are called Cevians. Altitudes, medians, angle bisectors are all Cevians and, in addition, conform to

Ceva's Theorem

Three Cevians AD, BE, and CF are concurrent iff

(1)

AF/FB · BD/DC · CE/EA = 1

holds.

For the mechanical proof, place a system of masses BF·CE, AF·CE, and BF·EA at the vertices A, B, and C, respectively. Then F becomes the center of mass of two points A and B, whereas E becomes the center of mass of points A and C. From here it follows that the center of mass of the three points lies on the intersection of Cevians BE and CF. (1) is then equivalent to the assertion that it also lies on the line AD. For details, see Ross Honsberger's Episodes or an online discussion.

I am grateful to Prof. William A. McWorter Jr for the following, I believe novel, proof and his advice on writing the Java demonstration below. The demonstration is limited to the case when all ratios in (1) are rational and the statement that the concurrency of the Cevians implies (1).

Create a grid of lines parallel to BE and CF. Refine the grid by dividing every segment parallel to BE into AF smaller segments. (To avoid introducing additional notations, AF also denotes the number of parts the segment AF is divided into, and the same is true of other segments involved.) Similarly, divide every segment parallel to CF into EA smaller parts. In the refined grid, AK (K being the common point of the three Cevians) serves as a diagonal of the "grid square" - parallelogram with equal number of units (EA·AF) on every side. Therefore, there is a set of diagonals parallel to AD that do not miss a single grid node.

FB is divided into FB·EA small segments by grid lines parallel to CF. Follow those grid lines towards BE and let them refract in BE to the direction of AD. We see that the grid diagonals divide BD into FB·EA small segments. Similarly (starting from CE), the diagonals divide DC into CE·AF parts. Because of the uniformity of the grid, BD/DC = FB·EA/CE·AF. Clearly,

(AF·EA / FB·EA) (FB·EA / CE·AF) (CE·AF / EA·AF) = 1

which is the same as

(AF/FB) (FB·EA / CE·AF) (CE/EA) = 1.

Thus (1) follows.

Mixing mechanical and geometric ideas, the proof is equivalent to placing masses AF·EA at A, FB·EA at B, and CE·AF at C, and then passing to the isotomic conjugates of the given Cevians. In strictly mechanical terms, we might, as well, place masses FB·EA·CE·AF, AF·EA·CE·AF, and FB·EA·AF·EA at vertices A, B, and C.

Cut an 8x8 chessboard into two triangles and two quadrilaterals. Then rearrange them as suggested by the right diagram. They will combine in a 5x13 rectangle. Well, it's easy to get fooled or fool somebody else if a real chessboard has been sacrificed for the sake of demonstration. A less expensive way is to use a piece of paper, a ruler and a soft pencil to draw "fat" lines. For an 8x8 puzzle, the computer, however, gives away the secret (but sometimes it does not). When the four pieces are arranged in a 5x13 rectangle a narrow parallelogram is left uncovered. (By Pick's theorem, the area of the parallelogram is exactly 1, for it contains no grid points in its interior, nor on its boundary, except for the four vertices.)

The numbers 5, 8, 13 are members of the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, ... We might have chosen another triple of successive Fibonacci numbers. For the triple 2, 3, 5, a less standard chessboard of 3×3 squares is transformed into a 2×5 rectangle. (In the applet, the only control parameter is the index of the smallest number in the triplet.)

In general, if Fk denotes the k-th Fibonacci number, a Fn×Fn square can be rearranged into a Fn-1×Fn+1 rectangle according to Cassini's identity that is easily proven by induction

(2)

Fn+1·Fn+1 - FnFn+2 = (-1)n

Regardless of whether we get surplus area or a deficit, the absolute error is always the same, 1. The relative error which is the ratio of the absolute error to Fn+12 decreases as n grows and becomes virtually indiscernible for n = 7 (in my resolution of 800x600.)

The goal of developing appreciation is often perceived as conflicting with the goal of acquiring knowledge. But to appreciate is to know in the first place. The real difference is in the ability to see a picture as a whole or only as a collection of its constituent parts. As with the bamboozlement puzzle, it's possible to see the polygons while missing the relationships that polygons stand to each other via their angles and side lengths.